Question

un muncitor are de executat 200 de piese.El executa zilnic cu 5 piese peste norma stabilită și termina lucrarea cu 2 zile mai repede. in câte zile a terminat lucrarea ?

Answer 1

Răspuns:

8 zile

Explicație pas cu pas:

Notam cu:

z - numarul de zile

n - numar de piese zilnic ( norma zilnica)

n * z = 200

executa cu 5 piese peste norma => n + 5 piese

termina cu 2 zile mai devreme => z -2 zile

(n + 5) * ( z - 2) = 200              (inmultim termenii si facem calculele)

$n*z - 2n +5 z - 5*2 = 200\\\\200 -2n+5z -10=200\\\\-2n +5z-10=0 ^{} ^{} ^{} / *z\\\\-2n*z +5 z^{2} -10z = 0\\\\-2*200 +5 z^{2} -10z = 0\\\\-400 + 5z^{2}-10z = 0 ^{} ^{} ^{} ^{} ^{} ^{} ^{} ^{} ^{} ^{} ^{} ^{} ^{} ^{} ^{} ^{}/:5 \\\\-80 + z^{2} - 2z = 0\\ \\z^{2} - 10z + 8z - 80 = 0\\ \\z *(z-10) + 8 * (z-10) =0$

$=> z-10 = 0 ^{} {} ^{} ^{} ^{} ^{} ^{} sau ^{} {} ^{} ^{} ^{} ^{} ^{} z + 8 = 0\\ \\z = 10 ^{} {} ^{} ^{} ^{} ^{} ^{} zile ^{} {} ^{} ^{} ^{} ^{} ^{} sau ^{} {} ^{} ^{} ^{} ^{} ^{} z = -8 ^{} {} ^{} ^{} ^{} ^{} ^{} nu ^{} {} ^{} ^{} ^{} ^{} ^{} convine\\\\n = 20 ^{} {} ^{} ^{} ^{} ^{} ^{} piese\\\\z -2 = 10 - 2 =>z = 8 ^{} {} ^{} ^{} ^{} ^{} ^{} ^{} {} ^{} ^{} ^{} ^{} ^{} zile$

Verificare: (20 +5) * 8 = 25 * 8 = 200

ai in atasament si varianta problemei dupa divizori, iar verificarea o faci prin povestea cu 2 zile